1 Res z k+1 (z c), 0 =

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1 32. COMPLEX ANALYSIS FOR APPLICATIONS Mock Final examination. (Monday June 7..am 2.pm) You may consult your handwritten notes, the book by Gamelin, and the solutions and handouts provided during the Quarter. No laptops or electronic calculators may be used. Maximum attainable points = 5. Each question carries bonus point awarded for a complete correct solution.. Each of the following functions has a singularity at =. Classify that singularity by determining whether it is isolated or not, and (if isolated) whether it is a branch point, an essential singularity or a pole. If it is an essential singularity or pole, find the residue. (a) sin, (b) 2 sin, (c) sin 2, (d) 4 sin. (9 points) 2. Show (for 4 points) that, if c is a constant and k is an integer, { c (k+) [, k,, k <, [ ] k+ ( c), = ] k+ ( c), c Use these results to obtain (for 5 points) the Laurent expansions of 2 ( 2 + )( + 2) for (a) <, (b) < < 2 and (c) > (a) Certain integrals of the form 2π f(sin θ, cos θ)dθ = c (k+). can be evaluated by complex variable methods. Explain very briefly the method. ( point) (b) Show that 2π cos(cos θ) cosh(sin θ) dθ = 2π, 2π sin(cos θ) sinh(sin θ) dθ =. ( points) 4. By integrating e ıa /( ) round a suitable contour, show (for 9 points) that, if a, ( + x 2 ) cos ax + x 2 + x 4 dx = π e 2 a 3 cos 3 2 a, x sin ax + x 2 + x 4 dx = π e 2 a 3 sin 3 2 a. 5. Describe very briefly what a keyhole contour is, and why it is useful. ( point) By integrating [Log /( + 2 )] 2 round a suitable keyhole contour, show (for 2 points) that log t ( + t 2 ) 2 dt = 4 π, and hence that t 2 log t ( + t 2 ) 2 dt = 4 π. (OVER

2 6. The sides of a pie-slice contour C are the segment x R of the positive real axis and the segment y R of the positive imaginary axis, the contour being completed by the arc θ 2 π of the circle = Reıθ. By integrating [e ı2 ]/ 2 round C, show (for 9 points) that [cos(x 2 ) + sin(x 2 ) ] dx x 2 =. 7. The contour C is a rectangle indented at = πı with sides y =, x = T >, y = π and x = S <. By integrating 2 / sinh round C and taking the limits S and T, show (for points) that x dx sinh x = 4 π2. 8. Show (for 9 points) that all five roots of = lie in the annulus < < 5 2 and that two lie in < < (a) Express f(x) = x 6 +4x 4 +x 3 +2x 2 +x+5 as a sum of quantities that are each non-negative for any real x. Hence show that f() = has no roots that are purely real. (5 points) (b) Find (for 6 points) the number of eros of f() = that lie in each quadrant. Sketch of Solutions. All four functions have isolated singularities at =. The nature of each is most easily determined by using Laurent expansions, which in the case of (a) (c) may be derived from the Taylor expansion sin = ( ) () (a) Let f() = / sin. By (), sin = 2 [ +...], so that f() = 2 [ +...]. Therefore = is a double pole. Since f() is even in [i.e., f( ) f()], the coefficient of in its expansion in is ero. Therefore [f(), ] =. (b) Let f() = / 2 sin. By (), 2 sin = 3 [ ], so that f() = 3 [ ]. Therefore = is a triple pole. The coefficient of in the expansion of f() in is 6. Therefore [f(), ] = 6. (c) Let f() = / sin 2. By (), sin 2 = 3 [ ] 2 = 3 [ ], so that f() = 3 [ ]. Therefore = is a triple pole. The coefficient of in its expansion in is 3. Therefore [f(), ] = 3. (d) Let f() = 4 sin. Here = is an essential singularity as can be seen from its Laurent expansion about = : f() = k= a k k, where a k = f()d 2πı =r k+ = Z k f(/z)dz (2) 2πı Z =r where Z = /. Since f(/z) = Z 4 sin Z = Z 3 [ 6 Z2 + 2 Z4 +...] for Z, the second form of a k in (2) shows that a k = for k 4. Since the expansion () is in powers of 2, we may infer that a 3 2m for m, so that the Laurent expansion of f() in negative powers of does not terminate, i.e., = is an essential singularity. The residue is the power of in (2), i.e. a = 2. 2

3 2. If k <, the function /[ k+ ( c)] does not have a singularity at =, so its residue at = is ero. If k, the function has a pole of order k + at =, and [ ] k+ ( c), = k! d k d k c = ( ) k k! = k! ( c) k+ = c (k+) = The simple pole at = c has a nonero residue for both k and k < ; it is [ ] k+ ( c), c = c (k+) Referring to (2) in Q. we confirm the binomial expansions c = k c k+, if < c; c = k= Use partial fractions to express the given function as k= k c k+ = c k k+ if > c (3) k= From (3) we have f() = = k= ( ) k 2 k+, if < 2; + 2 = k= ( 2) k k+ if > 2 and also 2 + = ( ) k 2k, if < ; k= 2 + = ( ) k 2(k+) if > k= Therefore the expansions in the three ranges of are (a) f() = (b) f() = (c) f() = ( ) k 2k+ + k= k= k= ( ) k 2k+ + ( ) k 2k+ + k= k= k= ( ) k+ 2 k+ ( ) k+ 2 k+ ( ) k+ 2 k k+ 3.(a) These integrals have been called Type in this course. They may (sometimes) be evaluated by converting them to integrals around the unit circle, by the transformation = e ıθ, dθ = d ı, cos θ = (eıθ + e ıθ ) = ( + ), sin θ = (eıθ e ıθ ) = ( ) 2 2 2ı 2ı (OVER 3

4 (b) cos(cos θ) cosh(sin θ) = 4 [e ı cos θ + e ı cos θ] [ e sin θ sin + e θ] Therefore, since cos θ + ı sin θ = and cos θ ı sin θ = / on =, we have where I = 4ı = 2π cos(cos θ) cosh(sin θ) dθ = I + I 2, [ e ı + e ı] d, I 2 = 4ı = [e ı/ + e ı/] d The integrand of I is analytic everywhere, apart from the simple pole at =, the residue at which is clearly 2, so that I = π. The integrand of I 2 is also analytic everywhere, apart from the essential singularity at =. We can obtain the value of this integral by picking out the coefficient of in the Laurent expansion of the exponentials at =. Or we can make the transformation /Z and convert I 2 into I 2 = 4ı Z = [e ız + e ız] dz Z which coincides with I, so that I 2 = π also. The final result is therefore 2π cos(cos θ) cosh(sin θ) dθ = 2π The argument is essentially the same for the other integral apart from minor changes. We find 2π sin(cos θ) sinh(sin θ) dθ = ı(i I 2 ) Since I = I 2 = π, we see that this integral is ero. 4. Consider C e ıa d, where C = A + S where A is the segment R R of the real axis and S is the semicircle = Re ıθ, θ π, R >. On S, e ıa = e ar sin θ e ıar cos θ = e ar sin θ. Also = ( 3 )/( ) (R 3 )/(R+). Thus, on S, the modulus of the integrand is bounded by M = (R+)/(R 3 ). Since the length of S is πr, it follows from the ML bound that e ıa S d πr(r + ) R 3 for R (4) The integrand has only one singularity in C, a simple pole at = 2 ( + ı 3). The residue at that pole is (using d Hospital s rule) [ e ıa ] + + 2, ( )e ıa = lim =eıa lim =eıa lim + 2 = ı 3 e 2 a( 3+ı) (5) 4

5 Equations (4) and (5) show that, in the limit R, The integral on the left-hand side is e ıax + x + x 2 dx + e ıax + x + x 2 dx = 2πı ı 3 e 2 a( 3+ı) (6) e ıax x + x 2 dx = where we have transformed the negative x part by x x. Therefore and (6) becomes e ıax + x + x 2 dx = 2 cos ax + ı sin ax cos ax ı sin ax + x + x 2 dx + x + x 2 dx ( + x 2 ) cos ax + x 2 + x 4 dx 2ı x sin ax + x 2 + x 4 dx ( + x 2 ) cos ax x sin ax 2π 2 + x 2 + x 4 dx 2ı + x 2 dx = e 2 a 3 (cos + x4 3 2 a ı sin 2 a) The real and imaginary parts of this give the stated results. 5. A keyhole contour is a way of eliminating a branch point from the interior of a contour C that would otherwise contain it. The key enters by a tiny, almost complete, circle surrounding the branch point and the two sides of the keyhole form the bridge connecting that circle to the rest of C. Because arg is different on the two sides of the bridge, the contributions made by these sides generally do not cancel, even though the two sides are described in opposite directions. Obviously the best place to locate the keyhole is on each side of the cut along the negative real axis required by Log. The contour C therefore consists of four parts: (i) S. A giant, almost complete, circle = Re ıθ, π < θ < π, R > ; (ii) s. A tiny, almost complete, circle, = re ıθ, π > θ > π, r < ; (iii) U. The upper side of the keyhole, on which = te ıπ where R t r; (iv) L. The lower side of the keyhole, on which = te ıπ where r t R. These make the following contributions: (i) On S, Log = ln R+ıθ, so that Log 2 = (ln R) 2 +θ 2 (ln R) 2 +π 2. Also 2 + R 2. An upper bound for the modulus of the integrand is therefore M = [(ln R) 2 +π 2 ]/(R 2 ) 2. The length of S is L = 2πR. The ML bound therefore gives S ( Log + 2 because (ln R) 2 /R 3 for R. ) 2 d 2πR[(ln R)2 + π 2 ] (R 2 ) 2, for R (OVER 5

6 (ii) On s, Log = ln r+ıθ, so that Log 2 = (ln r) 2 +θ 2 (ln r) 2 +π 2. Also 2 + r 2. An upper bound for the modulus of the integrand is therefore M = [(ln r) 2 +π 2 ]/( r 2 ) 2. The length of s is L = 2πr. The ML bound therefore gives s ( Log + 2 because r(ln r) 2 for r. (iii) On U, Log = ln t + ıπ and the integral is U ( ) Log 2 r [ln t + ıπ] d = R ( + t 2 ) 2 dt (iv) On L, Log = ln t ıπ and the integral is L ( ) Log 2 R [ln t ıπ] d = r ( + t 2 ) 2 dt Putting these results together, we see that S ) 2 d 2πr[(ln r)2 + π 2 ] ( r 2 ) 2, for r [ln t + ıπ] 2 ( + t 2 dt for r and R ) 2 [ln t ıπ] 2 ( + t 2 dt for r and R ) 2 ( ) Log 2 ln t + 2 d 4πı ( + t 2 dt, for r and R ) 2 The contour C contains two double poles, at = ı and = ı, and [ ( ) Log 2 + 2, ı] [ ( ) Log 2 + 2, ı] = lim ı = lim ı [ (Log ) ] 2 [ ] 2Log 2(Log )2 = + ı ( + ı) 2 ( + ı) 3 = 4 π + 6 π2 ı =ı ı [ (Log ) ] 2 [ ] 2Log 2(Log )2 = ı ( ı) 2 ( ı) 3 = 4 π 6 π2 ı = ı ı so that the sum of the residues is 2π. The stated result follows. The final integral is obtained by the transformation t /t. (dt dt/t 2, ln t ln t) 6. Since lim [e ı2 ]/ 2 = ı, there is no singularity of [e ı2 ]/ 2 at = or anywhere else in the finite plane, so that e ı2 2 d = (7) C For the integral along the quadrant S of the circle we have 2 = R 2 e 2ıθ and, since 2θ π on S, e ı2 = e R2 sin 2θ so that, by the triangle inequality, e ı2 2. The modulus of the integrand is therefore bounded by M = 2/R 2 and, since the length of S is L = 2πR, we have e ı2 2 d π R as R S The contributions to the integral from the sides of the pie along the x and y axes are 6

7 ı R R e ıx2 x 2 dx = e ıy2 y 2 dy = R R (cos x 2 + ı sin x 2 ) x 2 dx (ı cos x 2 + sin x 2 ı) x 2 dx respectively where, in the second integral, we have replaced the dummy integration variable y by x. Adding these together, we see that, in the limit R, From this and (7), the stated result follows. C e ı2 2 d = ( + ı) [cos x 2 + sin x 2 ] dx x 2 7. The eros of sinh are = mπı where m is an integer. These all lie outside C, apart from = and = πı, but the first of these is not a singularity of the integrand in I = C 2 d sinh because lim ( 2 / sinh ) =. The point = πı would, however, lie on C if we did not indent the contour there. We shall indent the contour below the line y = π so that the integral has no singularities in C, and therefore I =. On the right side R of the rectangle, = T + ıy, where y π, so that 2 = T 2 + y 2 T 2 + π 2. Also sinh = sinh T cos y + ı cosh T sin y, so that sinh = (sinh 2 T cos 2 y + cosh 2 T sin 2 y) = (sinh 2 T + sin 2 y) sinh T. The modulus of the integrand is therefore bounded by M = (T 2 + π 2 )/ sinh T and, since the length of R is π, the ML bound gives 2 d R sinh π(t 2 + π 2 ) as T sinh T Similarly, on the left side L of the rectangle, we have U 2 d L sinh π(s2 + π 2 ) as S sinh S Because of the indentation, the upper side, U, of the rectangle gives 2 d ɛ sinh = (x + ıπ) 2 S sinh(x + ıπ) s dx 2 d T sinh (x + ıπ) 2 dx (8) ɛ sinh(x + ıπ) where s denotes the small semicircle of the indentation, i.e., = πı + ɛe ıθ where π θ. The minus signs appear in (8) because C is described in the counterclockwise sense. In the limit ɛ, the integral on s tends to a half of what it would be for the complete circle πı = ɛ, i.e., πı times the residue of the simple pole at = πı. We have (OVER 7

8 [ ] 2 sinh, πı 2 = lim ( πı) πı sinh = πı π2 lim πı sinh = π2 lim πı cosh = π2 cos π = π2 In the limit ɛ, (8) therefore becomes 2 d T sinh = PV U S (x + ıπ) 2 sinh x dx π3 ı (9) where we have used the fact that sinh(x + πı) = sinh x. The integral in (9) is T S x 2 dx sinh x + 2πı T S x dx sinh x π2 PV T S dx sinh x Adding in the integral from the lower side of the rectangle (i.e., the x axis) we now have, in the limits T and S, x 2 dx I = 2 sinh x + 2πı x dx sinh x dx π2 PV sinh x π3 ı and, since I = (see above), we conclude that x dx sinh x = 2 π2, the first of which gives the stated result. 2x 2 π 2 PV sinh x dx = 8. This question is on the application of Rouché s theorem. Let f() = = 5 g() where g() = , g() = <, for = 5 2 Therefore, as f() describes = 5 2, g describes a curve α in the g plane lying inside the circle g =, so that α arg g =. Therefore =5/2 [arg f()] = 5 2π 5 roots within = 5 2. Let f() = 2 h() where h() = , h() = 23 < 2, for = 2. 2 Therefore, as f() describes = 2, h describes a curve β in the h plane lying inside the circle h+2 = 2, so that β arg h =. Therefore =2 [arg f()] = 2 2π 2 roots within = 2. Finally, f() = 3 < 4 on =, so that = [arg f()] = no roots within =. 9. This question is from G. VIII., #3, p Since every coefficient in f(x) is positive, it is completely clear that f(x) = does not have a positive root. The aim of the first part of the question is to establish that it does not have a negative root either. For general polynomials, this is non-trivial task, although part (b) shows in this case that there cannot be more than 2 negative roots. Here we can write f(x) as f(x) = x 6 + x 2 (2x + 4 ) x2 + (x + 2 ) () 8

9 The right-hand side is obviously positive for all real x so that f(x) = has no root that is purely real. The key to finding the form () is to get the terms odd in x (the only ones that could make f(x) negative) into squares. The expression () is not the only way of doing that; e.g., f(x) = (x 3 + x + 2 )2 + 2(x ) (b) Define a closed contour of 3 parts: (i) the real axis from = to = R ; (ii) the quadrant θ 2 π of the circle = Reıθ ; (iii) the imaginary axis from = ır to =. If you draw a diagram in the -plane, you might label these OA, AB and BO. From part (a), OA arg f() =, and it is fairly clear that AB arg f() 6 2π = 3π for R. The main task is to find BO arg f(), by tracking what f() does as y decreases from R( ) to on the imaginary axis. We have f(ıy) = y 6 + 4y 4 2y ıy(y 2 ) For y = R, both Re(f(ıy)) and Im(f(ıy)) are negative, and Re(f(ıR)) R 6 R 3 Im(f(ıR)). Thus, in the complex f() plane, the point starts just below the negative real axis. As y decreases, Re(f(ıy)) and Im(f(ıy)) remain negative until Re(f(ıy)) becomes positive at y 2 near 3.5. (The precise point where Re(f(ıy)) changes sign is unimportant. All one needs to establish is that, as y decreases, Re(f(ıy)) changes sign before Im(f(ıy)) does!) On decreasing y further, Re(f(ıy)) remains positive. Im(f(ıy)) changes sign at y and thereafter remains positive, until at y = where f(ıy) = 5. This helps in determining what argf() does in the complex f() plane as y decreases from R to. You will find that BO arg f() = π. Therefore, as does its tour OABO of the st quadrant of the plane, [arg[f()]] = + 3π + π = 4π. Therefore there are 2 roots in the st quadrant. Since the coefficients in f() are real, the roots occur in conjugate complex pairs. There are therefore 2 roots in the 4 th quadrant. Only 2 eros of the f() are unaccounted for, and by (a) they are not real and negative. Therefore is in the 2 nd quadrant and in the 3 rd quadrant. A note on grading: Hopefully you by now understand well the technique for evaluating the integrals that appear in Q4 7. The integral whose value is sought is along the real axis A. To make use of the power of Cauchy s theorem, you need to close the contour, by adding an integral around some curve S that takes you off the real axis and into the complex plane. But then you have to make sure that the integral that you ve added along S vanishes in the limit (or in the case of Q7 has a portion U that is closely related to the original integral along A). Look at the solution to Q.5 above, for example: ten lines are devoted to proving in (i) and (ii) that S and s make ero contributions in the limits R and r. Please do not ignore this important step by saying It can be shown that.... It is useless to leave it as an exercise for the examiner! If you do, you will lose points. Understand, and use the ML bound! 9